We proceed as in the previous question. First we take derivatives so that we can substitute into the DE.

\begin{align*}
y =\amp\ x^m \\
\Rightarrow \quad y' =\amp\ mx^{m-1} \\
\Rightarrow \quad y'' =\amp\ m(m-1)x^{m-2}
\end{align*}

Now we can substitute into the DE:

\begin{align*}
3x^2 \frac{d^2y}{dx^2} =\amp\
-11x \frac{dy}{dx} + 3y \\
3x^2\Big[ m(m-1)x^{m-2} \Big] =\amp\
-11x\Big[ mx^{m-1} \Big] + 3\Big[ x^m \Big] \\
3m(m-1)x^2 x^{m-2} =\amp\
-11mx^1\cdot x^{m-1} + 3x^m \\
(3m^2 - 3m)x^{2+(m-2)} =\amp\
-11m x^{1 + (m-1)} + 3x^m \\
(3m^2 - 3m)x^{m} =\amp\ -11m x^{m} + 3x^m \\
(3m^2 - 3m)x^{m} + 11m x^{m} - 3x^m =\amp\ 0 \\
(3m^2 - 3m + 11m - 3)x^m =\amp\ 0 \\
(3m^2 +8m - 3)x^m =\amp\ 0 \\
(3m -1)(m+3)x^m =\amp\ 0
\end{align*}

Note that there is no value of \(m\) that can guarantee that \(x^m\) is 0. Hence we have

\begin{align*}
3m-1 = 0 \amp \quad\text{or}\quad m+3 = 0 \\
m = \frac{1}{3}, \amp \quad\text{or}\quad m = -3
\end{align*}

So the two solutions are \(m = -3, \frac{1}{3}.\)